Question

is A,B and C are the interior angles of a triangle , prove that
sec A-B+C / 2 = cosec B​​

Answer 1

Step-by-step explanation:

Given, A,B and C are the interior angles of a triangle.

Therefore, A + B + C = 180

=> A + C = 180 - B

LHS :

Sec(A - B + C)/2

=> Sec(180 - B)/2

=> Sec(90 - B)

=> cosecB

Hope it helps!

Answer 2

HOPE IT HELPS YOU.........